A 16 Ω wire is bend to form a square loop. A 9 V supply having internal resistance of 1 Ω is connected across one of its sides. The potential drop across the diagonals of the square loop is × 10-1 V

Question

A 16 Ω wire is bend to form a square loop. A 9 V supply having internal resistance of 1 Ω is connected across one of its sides. The potential drop across the diagonals of the square loop is × 10-1 V (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

here assume current as <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9881511a9719751a44c57c16de209904-.png" /> By KVL in outer loop 9-12 i-4 i=0 16 i=9 8 i=(9/2)=4.5 =45 × 10-1

Q. A $16 Ω$ wire is bend to form a square loop. A $9 V$ supply having internal resistance of $1 Ω$ is connected across one of its sides. The potential drop across the diagonals of the square loop is_______ $\times 1 0^{- 1} V$

here assume current as

By KVL in outer loop
$9 - 12 i - 4 i = 0$
$16 i = 9$
$8 i = \frac{9}{2} = 4.5$
$= 45 \times 1 0^{- 1}$

Q. A 16Ω wire is bend to form a square loop. A 9V supply having internal resistance of 1Ω is connected across one of its sides. The potential drop across the diagonals of the square loop is_______ ×10−1V

here assume current as By KVL in outer loop 9−12i−4i=0 16i=9 8i=29​=4.5 =45×10−1

Q. A $16 Ω$ wire is bend to form a square loop. A $9 V$ supply having internal resistance of $1 Ω$ is connected across one of its sides. The potential drop across the diagonals of the square loop is_______ $\times 1 0^{- 1} V$

here assume current as

By KVL in outer loop
$9 - 12 i - 4 i = 0$
$16 i = 9$
$8 i = \frac{9}{2} = 4.5$
$= 45 \times 1 0^{- 1}$